Alathe is set to cut bars of steel into lengths of 6 centimeters. the lathe is considered to be in perfect adjustment if the average length of the bars it cuts is 6 centimeters. a sample of 144 bars is selected randomly and measured. it is determined that the average length of the bars in the sample is 6.09 centimeters with a standard deviation of .48 centimeters.

a. formulate the hypotheses to determine whether or not the lathe is in perfect adjustment.

b. compute the test statistic.

c. using the p-value approach, what is your conclusion? let ? = .05.

how do you compute using excel?

  We reject    H₀             ⇒   μ₀  = 6

Step-by-step explanation:

We are going to evaluate in a two tail test (the lathe cut above or below 6 ?)

So:

Normal Distribution with mean μ₀ = 6  and Standard deviation σ = 0.48

1.- Hipothesis test:

Null hypothesis                           H₀             ⇒   μ₀  = 6

Alternative hypothesis               Hₐ             ⇒   μ₀  ≠ 6

To compute the z(s)  we apply :

z(s) = [ ( μ  -   μ₀) ] / (σ√144)            ⇒  z(s) = [ ( 0.09 )* 12 ]/ .48

z(s) = 2.25

Now we have to compare z(s)   with  z(c) and we find z(c) from tables taken into account that   α = 0.05  or the confidence interval is 95 %

and as the test is a two tail one we get haft of α for each tail.

So we have to find the value of z(c) from an area value of 0.025

then  z(c) = 1.96

Then we have the situation in which z(s) > z(c)    2.25 > 1.96 that means we are in the rejection zone so we reject null hypothesis

Using excel we in any cell we do as follows

In a cell we insert fucntion and look for statistics and standard normal distribution

we introduce values of z(s) and we will get the probability of z(s) = .987 that probability is out of 0.95 the requierement of the problem so we reject       H₀             ⇒   μ₀  = 6

We also can use inverse normal distribution. In which case we must introduce areas (for instance 0,05) to find the z(c) and the compare

a. -0.3 > -0.33